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A random variable $K$ has hypergeometric distribution with parameters $N, m, n$, with probability mass function: $$ p_K(k)=\frac{\binom{m}{k}\binom{N-m}{n-k}}{\binom{N}{n}},\quad k\in\{\max(0,n+m-N),\ldots,\min(n,m)\}\quad, $$ $N$ is a population size, $m$ and $N-m$- sizes of two disjoint subsets of population (name them $A$ and $B$). The numerator indicates the number of $n$-element subsets, which contain $k$ elements choosen from $A$, and $n-k$ elements from $B$. I know the deriviation of this formula, however, i tried to do it in another way, and result is not correct. I would like to find out, where my reasoning fails. Suppose we firstly count a number of $n$-element sequences, in which we have on first $k$ places elements from $A$, and on the rest $n-k$ places elements of $B$. Then we divide this amount by a count of permutation to achieve figure of $n$-element subsets of population, with exactly $k$ elements from $A$ and $n-k$ elements of $B$: $$ \frac{\underbrace{m\times(m-1)\times\ldots\times(m-k+1)}_{k\ elements}\underbrace{(N-m)\times(N-m-1)\times\ldots\times(N-m-(n-k)+1)}_{n-k\ elements}}{n!}\\=\frac{m!}{(m-k)!}\frac{(N-m)!}{(N-m-(n-k))!}\frac{1}{n!}=\frac{\binom{m}{k}k!\binom{N-m}{n-k}(n-k)!}{n!}\ne \binom{m}{k}\binom{N-m}{n-k}\quad. $$ What is wrong with my derivation?

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Your numerator $$\frac{m!}{(m-k)!}\frac{(N-m)!}{(N-m-(n-k))!}$$ is the number of $n$-element sequences, in which we have on first $k$ places elements from $A$, and on the rest $n−k$ places elements of $B$. And in which the order matters, i.e. $(a_1,a_2,a_3,b_1,b_2)$ and $(a_1,a_3,a_2,b_2,b_1)$ are distinct $k+(n-k)$-element sequences.

You can't divide by $n!$ "to achieve figure of n-element subsets of population, with exactly k elements from A and n−k elements of B", because your sequence is splitted into two distinct subsequences, and you have $k!$ permutations in the first subsequence and $(n-k)!$ permutations in the second one. To count the $n$-element subsets with exactly $k$ elements from $A$, followed by $(n-k)$ elements from $B$, you must divide by $k!(n-k)!$.

A simple example. Let's say that $n=4$ and $k=2$. A single sequence may be: $$a_1,a_2,b_1,b_2$$ There are $n!=4!=24$ permutations, e.g. $$a_1,a_2,b_1,b_2;\quad a_2,a_1,b_1,b_2;\quad a_1,a_2,b_2,b_1\quad a_2,a_1,b_2,b_1\tag{1}$$ but also: $$b_1,a_2,b_2,a_1;\quad b_1,b_2,a_1,a_2;\quad etc.\tag{2}$$ In (2) you do not have $k=2$ elements of $A$ followed by $n-k=2$ elements of B, as you have in (1), and the number of permutations in (1) is just $k!(n-k)!=2\times 2=4$: the number of permutations of $k$ elements of $A$, multiplied by the number of permutations of the following $n-k$ elements of $B$.

When you divide by $k!(n-k)!$ you get the number of $n$-elements subsets where the order does not matter: $$\frac{m!}{(m-k)!k!}\frac{(N-m)!}{(N-m-(n-k))!(n-k)!}=\binom{m}{k}\binom{N-m}{n-k}$$

But this is still the "number of favorable cases", where $k$ is fixed. To get its probability you have to divide it by the "number of all cases possibile", which is $\binom{N}{n}$.

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    $\begingroup$ Some problems with your answer, e.g., ${m\choose k}\ne m(m-1)\cdots (m-k+1)$. $\endgroup$ – Dedekind Cuts Jul 19 at 15:57
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    $\begingroup$ Ops! I was a bit hasty :) $\endgroup$ – Sergio Jul 19 at 16:19
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    $\begingroup$ @DedekindCuts Thanks! $\endgroup$ – Sergio Jul 19 at 17:06
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    $\begingroup$ @Mentossinho Ok. I'll edit my answer. $\endgroup$ – Sergio Jul 19 at 17:49
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    $\begingroup$ Sorry: the amount of all kind of $n$-sequences is $N!/(N-n)!$ if the order matters. $\endgroup$ – Sergio Jul 19 at 20:07

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